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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Proof:</dfn> Substituting <span class="process-math">\(Y_1\)</span> and <span class="process-math">\(Y_2\)</span> into (<a href="" class="xref" data-knowl="./knowl/eq3_11.html" title="Equation 3.6.1">(3.6.1)</a>), one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_11.html ./knowl/eq3_12.html">
\begin{equation}
\begin{aligned}
&amp;Y_1^{\prime \prime}+p(x) Y_1^{\prime}+q(x) Y_1=g(x),\\
&amp;Y_2^{\prime \prime}+p(x) Y_2^{\prime}+q(x) Y_2=g(x).
\end{aligned}\tag{3.6.3}
\end{equation}
</div>
<p class="continuation">Then <span class="process-math">\((\ref{eq3_13})_2-(\ref{eq3_13})_1\)</span> gives:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_11.html ./knowl/eq3_12.html">
\begin{equation*}
(Y_2-Y_1)^{\prime \prime}+p(x) (Y_2-Y_1)^{\prime}+q(x)(Y_2-Y_1)=0,
\end{equation*}
</div>
<p class="continuation">which implies <span class="process-math">\(Y_2-Y_1\)</span> satisfies (<a href="" class="xref" data-knowl="./knowl/eq3_12.html" title="Equation 3.6.2">(3.6.2)</a>).</p>
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